No, it is not possible
Yes, it is possible

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Define the sequence $a_n$ by the following:

$a_n=\frac{k}{N-n}, \ 1 \leq n<N$

where $k$ and $N$ are fixed positive integers.

We'll prove that for some appropiate $k$ and $N$ this sequence fulfills the problem's conditions.

First, we seek to show that the difference $\frac{1}{a_i}-\frac{1}{a_{i+1}}$ does not depend on the value of $i$ . Indeed,

$\frac{1}{a_i}-\frac{1}{a_{i+1}} = \frac{N-i}{k}-\frac{N-i-1}{k} = \frac{1}{k}$ .

Letting $k=N!$ it follows that ${N-n} | k$ for every $1 \leq n<N$ , hence the value of $a_n=\frac{k}{N-n}$ is a positive integer. Since the condition $a_i<a_{i+1}$ is the simple consequence of the fact that $\frac{k}{N-i} < \frac{k}{N-i-1}$ , we have shown that this sequence is a correct solution for any $N$ . Putting in $N=2018$ we are done.